Can C++ lambda-expression throw? - c++

I can't find anything about throwing exceptions during constructing closure object.
It's oblivious that this expression can throw during copy construction of the vector:
auto v = std::vector<int>(1000000);
[v]{};
But what about empty or "by reference" capture lists like this:
[&]{};
I'm speaking only about construction of closure object now. Calling isn't interesting.
I read 5.1.2 Lambda expressions [expr.prim.lambda], but found nothing special about nothrow guarantee.

According to the standard (draft n3337), §5.1.2/3:
The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion
class type — called the closure type — whose properties are described below. This class type is not
an aggregate (8.5.1). The closure type is declared in the smallest block scope, class scope, or namespace
scope that contains the corresponding lambda-expression.
In a nutshell, a lambda expression ultimately instantiates an anonymous type (known only to the compiler), containing member variables that store the values indicated in the capture list. So imagine you saw a class that looked like this:
class __IgnoreMe__
{
std::vector<int>& _v;
public:
__IgnoreMe__(std::vector<int>& v)
: _v(v)
{
}
void operator()()
{
}
};
(Note: this isn't exactly how the class created by the compiler looks. The standard lays out specific requirements for generated classes, which I've left out for brevity's sake.)
Now imagine you instantiate that class like this:
auto v = std::vector<int>(1000000);
auto my_lambda = __IgnoreMe__(v);
Can the instantiation of my_lambda throw an exception? If so, then the construction of the closure object can throw an exception. (It can't, in this case.)
As for providing nothrow guarantees, the standard doesn't require compilers to account for this, but it doesn't prevent them from doing it either. The end of §5.1.2/3 states:
An implementation may define the closure
type differently from what is described below provided this does not alter the observable behavior of the
program other than by changing:
— the size and/or alignment of the closure type,
— whether the closure type is trivially copyable (Clause 9),
— whether the closure type is a standard-layout class (Clause 9), or
— whether the closure type is a POD class (Clause 9).

Related

Is it valid to create closure (lambda) objects using `std::bit_cast` in C++20?

A colleague showed me a C++20 program where a closure object is virtually created using std::bit_cast from the value that it captures:
#include <bit>
#include <iostream>
class A {
int v;
public:
A(int u) : v(u) {}
auto getter() const {
if ( v > 0 ) throw 0;
return [this](){ return v; };
}
};
int main() {
A x(42);
auto xgetter = std::bit_cast<decltype(x.getter())>(&x);
std::cout << xgetter();
}
Here main function cannot call x.getter() due to exception. Instead it calls std::bit_cast taking as template argument the closure type decltype(x.getter()) and as ordinary argument the pointer &x being captured for new closure object xgetter. Then xgetter is called to obtain the value of object x, which is otherwise not accessible in main.
The program is accepted by all compilers without any warnings and prints 42, demo: https://gcc.godbolt.org/z/a479689Wa
But is the program well-formed according to the standard and is such 'construction' of lambda objects valid?
But is the program well-formed according to the standard ...
The program has undefined behaviour conditional on leeway given to implementors. Particularly conditional on whether the closure type of the lambda
[this](){ return v; };
is trivially copyable from; as per [expr.prim.lambda.closure]/2:
The closure type is declared in the smallest block scope, class scope,
or namespace scope that contains the corresponding lambda-expression. [...]
The closure type is not an aggregate type. An
implementation may define the closure type differently from what is
described below provided this does not alter the observable behavior
of the program other than by changing:
(2.1) the size and/or alignment of the closure type,
(2.2) whether the closure type is trivially copyable ([class.prop]), or
(2.3) whether the closure type is a standard-layout class ([class.prop]). [...]
This means that whether the constraints of [bit.cast]/1 are fulfilled or not:
template<class To, class From>
constexpr To bit_cast(const From& from) noexcept;
Constraints:
(1.1) sizeof(To) == sizeof(From) is true;
(1.2) is_­trivially_­copyable_­v<To> is true; and
(1.3) is_­trivially_­copyable_­v<From> is true.
is implementation-defined.
... and is such 'construction' of lambda objects valid?
As [expr.prim.lambda.closure]/2.1 also states that the size and alignment of the closure type is implementation-defined, using std::bit_cast to create an instance of the closure type may result in a program with undefined behavior, as per [bit.cast]/2:
Returns: An object of type To. Implicitly creates objects nested within the result ([intro.object]). Each bit of the value representation of the result is equal to the corresponding bit in the object representation of from. Padding bits of the result are unspecified. For the result and each object created within it, if there is no value of the object's type corresponding to the value representation produced, the behavior is undefined. If there are multiple such values, which value is produced is unspecified.
For any kind of practical use, however, I'd argue that if a construct has undefined behavior conditional on implementation leeway details (unless these can be queried with say traits), then the construct should reasonably be considered to have undefined behavior, except possibly for a compiler's internal C++ (e.g. Clang frontend) implementation, where these implementation details are known.

Are captureless lambda guaranteed to be empty by the standard?

I'm searching for a way to identify empty (captureless) lambdas from other lambdas in a template function. I'm currently using C++17 but I'm curious for C++20 answers too.
My code looks like this:
template<typename T>
auto func(T lambda) {
// The aguments of the lambdas are unknown
if constexpr (/* is captureless */) {
// do stuff
}
}
Is it guaranteed by the C++ standard (17 or 20) that a captureless lambda, which is convertible to a function pointer, will also make std::is_empty yield true?
Take this code as an example:
auto a = []{}; // captureless
auto b = [c = 'z']{}; // has captures
static_assert(sizeof(a) == sizeof(b)); // Both are the same size
static_assert(!std::is_empty_v<decltype(b)>); // It has a `c` member
static_assert(std::is_empty_v<decltype(a)>); // Passes. It is guaranteed?
Live example
No, in fact, the standard explicitly grants permission for lambdas to have a size that doesn't line up with their declaration. [expr.prim.lambda.closure]/2 states
The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression. [ Note: This determines the set of namespaces and classes associated with the closure type ([basic.lookup.argdep]). The parameter types of a lambda-declarator do not affect these associated namespaces and classes. — end note ] The closure type is not an aggregate type. An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program other than by changing:
the size and/or alignment of the closure type,
whether the closure type is trivially copyable ([class.prop]), or
(2.3)
whether the closure type is a standard-layout class ([class.prop]).
An implementation shall not add members of rvalue reference type to the closure type.
emphasis mine
So this allows the implementation to give the lambda a member even if it is capture-less. I don't think any implementation ever would, but they are legally allowed to do so.

When is a lambda trivial?

When is a lambda a guaranteed to be trivial, if ever?
I assumed that if it captures only trivial types or nothing, it would be trivial. I don't have any standard-ese to back that up though.
My motivation was in moving some code from Visual C++ 12 to 14 and discovered some static asserts failed when dealing with lambdas I assumed to be trivial.
Example:
#include <type_traits>
#include <iostream>
using namespace std;
int main()
{
auto lambda = [](){};
cout << boolalpha << is_trivially_copyable<decltype(lambda)>{} << endl;
}
This produces false on vs140 but true in vs120 and clang. I could not test gcc due to not having gcc >= 5 around. I expect this is a regression in vs140, but I'm not certain of the correct behavior here.
The standard does not specify whether a closure type (the type of a lambda expression) is trivial or not. It explicitly leaves this up to implementation, which makes it non-portable. I am afraid you cannot rely on your static_assert producing anything consistent.
Quoting C++14 (N4140) 5.1.2/3:
... An implementation may define the closure type differently from what is described below provided this does not alter the observable
behavior of the program other than by changing:
the size and/or alignment of the closure type,
whether the closure type is trivially copyable (Clause 9),
whether the closure type is a standard-layout class (Clause 9), or
whether the closure type is a POD class (Clause 9).
...
(Emphasis mine)
After parsing the double negation in that sentence, we can see that the implementation is allowed to decide whether the closure type is trivially copyable, standard layout, or POD.
Note that the same wording is also present in C++17 (N4659), [expr.prim.lambda.closure] 8.1.5.1/2.
According to the draft standard N4527 5.1.2/3 Lambda expressions [expr.prim.lambda] (emphasis mine):
The type of the lambda-expression (which is also the type of the
closure object) is a unique, unnamed nonunion class type — called the
closure type — whose properties are described below. This class type
is neither an aggregate (8.5.1) nor a literal type (3.9). The closure
type is declared in the smallest block scope, class scope, or
namespace scope that contains the corresponding lambda-expression. [
Note: This determines the set of namespaces and classes associated
with the closure type (3.4.2). The parameter types of a
lambdadeclarator do not affect these associated namespaces and
classes. — end note ] An implementation may define the closure type
differently from what is described below provided this does not alter
the observable behavior of the program other than by changing:
(3.1) — the size and/or alignment of the closure type,
(3.2) — whether the closure type is trivially copyable (Clause 9),
(3.3) — whether the
closure type is a standard-layout class (Clause 9), or
(3.4) — whether
the closure type is a POD class (Clause 9).
An implementation shall
not add members of rvalue reference type to the closure type
Thus, it's implementation dependent.

C++11 lambdas can access my private members. Why?

Consider this piece of code:
class shy {
private:
int dont_touch; // Private member
public:
static const shy object;
};
const shy shy::object = []{
shy obj;
obj.dont_touch = 42; // Accessing a private member; compiles; WHY?
return obj;
}();
int main()
{
}
Live code (Clang)
Live code (GCC)
It seems really unintuitive to me. What does the C++11/14 standard say about this? Is this a GCC/Clang bug?
As already answered in the comments #Tony D and #dyp:
§ 9.4.2/2 Static data members [class.static.data]:
The initializer expression in the
definition of a static data member is in the scope of its class.
The above means that static data members and their initializers can access other private and protected members of their class.
Also consdering the standard § 5.1.2/2&3 Lambda expressions [expr.prim.lambda]:
2 The evaluation of a lambda-expression results in a prvalue temporary (12.2). This temporary is called the closure object. A lambda-expression shall not appear in an unevaluated operand (Clause 5). [ Note: A closure object behaves like a function object (20.9).-end note]
3 The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion class type - called the closure type - whose properties are described below. This class type is not an aggregate (8.5.1). The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression.
Combining the above we end up to the conclusion that the prvalue temporary closure object of your lambda is declared and defined in the initializer of the static const data member shy::object and consequently the scope of the lambda's closure object is the scope of class shy. As such it can access private members of class shy.

Type of lambda function [duplicate]

This question already has answers here:
What is the type of lambda when deduced with "auto" in C++11?
(8 answers)
Closed 6 years ago.
There is this code:
auto fun = [](int x)->int {return x + 1; };
std::cout << typeid(fun).name() << std::endl;
The result is: Z4mainEUliE_ but c++filt doesn't seem to explain what is it. What is type of lambda expression?
§5.1.2/3 states:
The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type
It goes on to say more, but that's the most important bit. A lambda is basically an instance of an anonymous class.
Incidentally, the demangled form of your lambda is main::$_0.
The type of a lambda function is unspecified by the standard (§5.1.2):
The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union classtype — called the closure type — whose properties are described below. This class type is not an aggregate (8.5.1). The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression.
It then goes on listing the exact properties a closure type should have.
Therefore there is no general type for a lambda function to have. The compiler will generate a new functor type with unspecified name for each lambda function
What is type of lambda expression?
The type of a lambda expression (the so-called closure) is an unnamed class type with a function call operator automatically generated by the compiler. The internal name the compiler will give it is unspecified.
According to Paragraph 5.1.2/3 of the C++11 Standard:
The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion
class type — called the closure type — whose properties are described below. This class type is not
an aggregate (8.5.1). The closure type is declared in the smallest block scope, class scope, or namespace
scope that contains the corresponding lambda-expression. [...]
Also notice, that the name() member function of the type_info class (the type returned by typeid()) is also implementation-dependent, and the Standard does not require it to be meaningful for a human.
Per Paragraph 18.7.1:
const char* name() const noexcept;
9 Returns: An implementation-defined NTBS.
10 Remarks: The message may be a null-terminated multibyte string (17.5.2.1.4.2), suitable for conversion and display as a wstring (21.3, 22.4.1.4)